Chapter 15: Problem 2183
In electromagnetic spectrum, the visible light lie between (A) radiowaves and microwaves (B) ultraviolet rays and infrared rays (C) ultraviolet rays and \(\mathrm{x}\) rays (D) infrared rays and microwaves
Chapter 15: Problem 2183
In electromagnetic spectrum, the visible light lie between (A) radiowaves and microwaves (B) ultraviolet rays and infrared rays (C) ultraviolet rays and \(\mathrm{x}\) rays (D) infrared rays and microwaves
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Get started for freeRange of frequency of microwaves is about (A) \(530 \mathrm{kHz}\) to \(1710 \mathrm{kHz}\) (B) \(54 \mathrm{MHz}\) to \(890 \mathrm{MHz}\) (C) \(3 \mathrm{GHz}\) to \(300 \mathrm{GHz}\) (D) \(4 \times 10^{14} \mathrm{~Hz}\) to \(7 \times 10^{14} \mathrm{~Hz}\)
When an electromagnetic wave encounters a dielectric medium, the transmitted wave has (A) same frequency but different amplitude (B) same amplitude but different frequency (C) same frequency and amplitude (D) different frequency and amplitude
A plane electromagnetic wave is incident on a mater1al surface. The wave delivers momentum \(P\) and energy \(E\) (A) \(\mathrm{P}=0, \mathrm{E} \neq 0\) (B) \(\mathrm{P} \neq 0, \mathrm{E}=0\) (C) \(P \neq 0, E \neq 0\) (D) \(P=0, E=0\)
The waves used in communication are generally called (A) \(\gamma\) rays (B) \(\alpha\) rays (C) microwaves (D) radiowaves
Astronomers have found that electromagnetic waves of wavelength $21 \mathrm{~cm}$ are continuously reaching the Earth's surface. Calculate the frequency of this radiation. $\left(\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ (A) \(14.28 \mathrm{GHz}\) (B) \(1.428 \mathrm{kHz}\) (C) \(1.428 \mathrm{MHz}\) (D) \(1.428 \mathrm{GHz}\)
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